It is given that circumference of $C$ is $l$ and the perimeter of triangle T is $l .$

Now, let the radius of circle C is r, so

$2\pi r=l\Rightarrow r=\frac{l}{2\pi }$

$\therefore$ area of circle C is $A_1=\pi r^2=\frac{l^2}{4\pi }$

Now, as we know that area of triangle will be maximum for given perimeter if it is an equilateral triangle, let the length of side of equilateral triangle is ' a, then

$3a=l\Rightarrow a=\frac{l}{3}$

and area of equilateral triangle is

$A_2=\frac{\sqrt{3}}{4}{a}^2$ 

$\mathrm{So, }{A}_2=\frac{\sqrt{3}}{4}\left(\frac{l^2}{9}\right)=\frac{l^2}{12\sqrt{3}}$ $\because \frac{A_1}{A_2}=\frac{\frac{l^2}{4\pi }}{\frac{l^2}{12\sqrt{3}}}=\frac{3\sqrt{3}}{\pi }>1$

since, as we took an equilateral triangle, which has maximum area. But we can take a triangle T such that the ratio $\frac{\text{ area }\left(C\right)}{\text{ area }\left(T\right)}$ is greater than any positive real number $\alpha$.